Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the gradient \(\nabla f\). $$ f(x, y)=x^{2} y /(x+y) $$

Step-by-Step Solution

Answer

The gradient is \( \nabla f = \left( \frac{x^2 y + 2xy^2}{(x + y)^2}, \frac{x^3}{(x + y)^2} \right) \).

Step 1: Understand the Gradient

The gradient of a multivariable function \( f(x, y) \), denoted as \( abla f \), is a vector consisting of partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). We will compute each of these partial derivatives.

Step 2: Differentiate with Respect to x

Find \( \frac{\partial f}{\partial x} \). Using the function \( f(x, y) = \frac{x^2 y}{x + y} \), apply the quotient rule: \(abla f\). Quotient rule: \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \( u = x^2 y \) and \( v = x + y \). Compute \( u' = 2xy \) and \( v' = 1 \).\So, \( \frac{\partial f}{\partial x} = \frac{ (x + y)(2xy) - (x^2 y)(1)}{(x + y)^2} = \frac{2x^2 y + 2xy^2 - x^2 y}{(x + y)^2} = \frac{x^2 y + 2xy^2}{(x + y)^2} \).

Step 3: Differentiate with Respect to y

Find \( \frac{\partial f}{\partial y} \). Again, use the quotient rule for \( f(x, y) = \frac{x^2 y}{x + y} \). \( u = x^2 y \) and \( v = x + y \). Differentiate: \( u' = x^2 \) and \( v' = 1 \).\Thus \( \frac{\partial f}{\partial y} = \frac{ (x + y)(x^2) - (x^2 y)(1)}{(x + y)^2} = \frac{x^3 + x^2 y - x^2 y}{(x + y)^2} = \frac{x^3}{(x + y)^2} \).

Step 4: Combine the Partial Derivatives

Combine both partial derivatives to find the gradient: \( abla f = \left( \frac{x^2 y + 2xy^2}{(x + y)^2}, \frac{x^3}{(x + y)^2} \right) \). This vector represents the gradient of \( f(x, y) \).

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as we change one variable while keeping others constant. For instance, if we have a function of two variables, such as the one given, \( f(x, y) = \frac{x^2 y}{x+y} \), partial derivatives measure how \( f \) changes with respect to \( x \) and \( y \) separately.
To find the partial derivative of a function with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), we treat \( y \) as a constant and differentiate \( f \) with respect to \( x \). Conversely, to find \( \frac{\partial f}{\partial y} \), \( x \) is held constant. Partial derivatives play a crucial role in finding gradients of functions, which, in turn, are useful in understanding how the function behaves in a multidimensional space.

Here, calculating \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) involves using rules like the quotient rule to properly differentiate the function.

Quotient Rule

The quotient rule is a fundamental technique in calculus used to differentiate functions that are expressed as a division of two other functions, like \( f(x, y) = \frac{u}{v} \). When working with the partial derivatives of \( f(x, y) = \frac{x^2 y}{x + y} \), the quotient rule becomes indispensable.
The quotient rule formula is:

• \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)

For our function, \( u = x^2 y \) and \( v = x + y \). Consequently, when calculating \( \frac{\partial}{\partial x} \) or \( \frac{\partial}{\partial y} \), we substitute \( u \) and \( v \) into the quotient rule, and perform the necessary differentiation.
This involves computing the derivatives \( u' \) and \( v' \), and substituting into the formula, effectively managing both the numerator and the denominator. The quotient rule thus ensures we correctly account for both functions' changes when deriving their expression.

Multivariable Function

A multivariable function involves more than one variable, such as \( f(x, y) = \frac{x^2 y}{x + y} \). These functions are more complex because changes in their values depend on variations in multiple variables.
Understanding multivariable functions is essential for analyzing relationships where two or more factors interact to determine outcomes, like in physics or economics.
When working with multivariable functions, tools like partial derivatives and the gradient are essential as they tell us how the function behaves when variables change independently. This sophisticated analysis allows us to capture the behavior of complex systems, paving the way for further exploration in fields like data science.
It’s also visually depicted as a surface in a 3D space, where each point \((x, y, f(x, y))\) exhibits how the function reacts to varying values.

Vector Calculus

Vector calculus extends regular calculus into multiple dimensions, providing tools to analyze vector fields and functions of several variables. The gradient, \( abla f \), is a vector itself that points in the direction of greatest increase of the function. In this context, it uses partial derivatives, such as those calculated for \( f(x, y) \) in the example given.
With our function, \( abla f = \left( \frac{x^2 y + 2xy^2}{(x + y)^2}, \frac{x^3}{(x + y)^2} \right) \), capturing both the rate and the direction of \( f \)'s steepest ascent.

This gradient vector is a core concept in vector calculus, often used in areas like physics for understanding fields such as electromagnetism, where direction and magnitude are both crucial. By applying vector calculus, we not only find the change but also the best-fitting line or path, enabling robust modeling and prediction of complex phenomena.